Question

What is the minimum value of 3x + 5y in the feasible region? (0, 7), (3, 7), (6, 3), (6, 0)

Answer 1

Answer:

The minimum value is $18$

It occurs at $P(6,0)$

Step-by-step explanation:

The given objective function is $P(x,y)=3x+5y$

The vertices of the feasible region are $(0,7),(3,7),(6,3),(6,0)$.

We plug in the vertices in to feasible region and evaluate to obtain,

$P(0,7)=3(0)+5(7)$

$P(0,7)=0+35$

$P(0,7)=35$

$P(3,7)=3(3)+5(7)$

$P(3,7)=9+35$

$P(3,7)=44$

$P(6,3)=3(6)+5(3)$

$P(6,3)=18+15$

$P(6,3)=33$

$P(6,0)=3(6)+5(0)$

$P(6,0)=18+0$

$P(6,0)=18$

Answer 2

Answer:

The minimum value is $18$

It occurs at $P(6,0)$

Step-by-step explanation:

The given objective function is $P(x,y)=3x+5y$

The vertices of the feasible region are $(0,7),(3,7),(6,3),(6,0)$.

We plug in the vertices in to feasible region and evaluate to obtain,

$P(0,7)=3(0)+5(7)$

$P(0,7)=0+35$

$P(0,7)=35$

$P(3,7)=3(3)+5(7)$

$P(3,7)=9+35$

$P(3,7)=44$

$P(6,3)=3(6)+5(3)$

$P(6,3)=18+15$

$P(6,3)=33$

$P(6,0)=3(6)+5(0)$

$P(6,0)=18+0$

$P(6,0)=18$